M.C.Q (1 Marks)

Question 511 Mark

Evaluate : $\int \frac{x^3}{x+2} d x$

Answer

(b) : Let $I=\int \frac{x^3}{x+2} d x$
Dividing $x^3$ by $x+2$, we get
$
=\int\left(x^2-2 x+4-\frac{8}{x+2}\right) d x=\frac{x^3}{3}-x^2+4 x-8 \log |x+2|+C
$

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Question 521 Mark

Evaluate: $\int \sec ^2(7-4 x) d x$

Answer

(d) : Let $I=\int \sec ^2(7-4 x) d x$
Put $7-4 x=t \Rightarrow d x=\frac{-1}{4} d t$
$
\therefore \quad I=\int \frac{\sec ^2 t}{-4} d t \Rightarrow I=\frac{\tan t}{-4}+C=\frac{\tan (7-4 x)}{-4}+C
$

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Question 531 Mark

Evaluate: $\int[\sin (\log x)+\cos (\log x)] d x$

Answer

(a) : Let $I=\int[\sin (\log x)+\cos (\log x)] d x$
Put $\log x=t \Rightarrow x=e^t \quad \Rightarrow d x=e^t d t$
$
\begin{aligned}
\therefore \quad I & =\int(\sin t+\cos t) e^t d t=e^t \sin t+C \\
& =x \sin (\log x)+C\left[\because\left[f(x)+f^{\prime}(x)\right] e^x d x=e^x f(x)+C\right)
\end{aligned}
$

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Question 541 Mark

Evaluate: $\int \frac{\left(a^x+b^x\right)^2}{a^x b^x} d x$

Answer

(a) : We have, $\int \frac{\left(a^x+b^x\right)^2}{a^x b^x} d x=\int \frac{a^{2 x}+b^{2 x}+2 a^x b^x}{a^x b^x} d x$
$
=\int\left(\left(\frac{a}{b}\right)^x+\left(\frac{b}{a}\right)^x+2\right) d x=\frac{\left(\frac{a}{b}\right)^x}{\log \frac{a}{b}}+\frac{\left(\frac{b}{a}\right)^x}{\log \frac{b}{a}}+2 x+C, a \neq b
$

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Question 551 Mark

Evaluate: $\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x$

Answer

$(d) :$ We have, $\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x$
Integrating by parts, we get
$\int_2^4 \frac{\left(x^2+x\right)}{\sqrt{2 x+1}} d x=\left[\left(x^2+x\right) \cdot \sqrt{2 x+1}\right]_2^4-\int_2^4(2 x+1) \cdot \sqrt{2 x+1} d x$
$=(60-6 \sqrt{5})-\int_2^4(2 x+1)^{3 / 2} d x$
$=(60-6 \sqrt{5})-\frac{1}{5} \cdot\left[(2 x+1)^{5 / 2}\right]_2^4$
$=(60-6 \sqrt{5})-\left(\frac{243}{5}-5 \sqrt{5}\right)=\left(\frac{57}{5}-\sqrt{5}\right)=\left(\frac{57-5 \sqrt{5}}{5}\right)$

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Question 561 Mark

Evaluate: $\int \frac{\sec ^2 x}{2+\tan x} d x$

Answer

(c) : Let $I=\int \frac{\sec ^2 x}{2+\tan x} d x$
Put $2+\tan x=t \Rightarrow \sec ^2 x d x=d t$
$
\therefore \quad I=\int \frac{d t}{t}=\log |t|+C=\log |2+\tan x|+C
$

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Question 571 Mark

Evaluate: $\int \frac{d x}{\sqrt{1-2 x-x^2}}$

Answer

$\text { (c) : Let } I=\int \frac{d x}{\sqrt{1-\left(x^2+2 x\right)}}=\int \frac{d x}{\sqrt{2-\left(x^2+2 x+1\right)}}$
$=\int \frac{d x}{\sqrt{2-(1+x)^2}}=\int \frac{d x}{\sqrt{(\sqrt{2})^2-(1+x)^2}}$
Put $1+x=z \Rightarrow d x=d z$
$\therefore I=\int \frac{d z}{\sqrt{(\sqrt{2})^2-z^2}}=\sin ^{-1} \frac{z}{\sqrt{2}}+C=\sin ^{-1}\left(\frac{1+x}{\sqrt{2}}\right)+C$

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Question 581 Mark

Evaluate: $\int_0^{\pi / 4} \tan ^3 x d x$

Answer

Let $I= \int_0^{\pi / 4} \tan ^3 x d x=\int_0^{\pi / 4}\left(\sec ^2 x-1\right) \tan x d x$
$ =\int_0^{\pi / 4} \sec ^2 x \tan x d x-\int_0^{\pi / 4} \tan x d x$
Put $\tan x=t$ in first integral
$\Rightarrow \sec ^2 x d x=d t$
When
$x=0 $
$\Rightarrow t=0$
$\therefore x=\pi / 4 $
$\Rightarrow t=1$
$I=\int_0^1 t d t-\int_0^{\pi / 4} \tan x d x$
$=\left[\frac{t^2}{2}\right]_0^1-[\log |\sec x|]_0^{\pi / 4}$
$=\left(\frac{1}{2}-0\right)-\log \left|\sec \frac{\pi}{4}\right|+\log |\sec 0|$
$=\frac{1}{2}(1-\log 2)$

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Question 591 Mark

Evaluate: $\int 2^{2^{2^x}} 2^{2^x} 2^x d x$

Answer

(a) : Let $I=\int 2^{2^{2^x}} 2^{2^x} 2^x d x$
Let $2^{2^{2^x}}=t \Rightarrow 2^{2^{2^x}} 2^{2^x} 2^x(\log 2)^3 d x=d t$
$
\Rightarrow I=\int \frac{1}{(\log 2)^3} d t=\frac{1}{(\log 2)^3} t+C=\frac{1}{(\log 2)^3} 2^{2^{2^x}}+C
$

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Question 601 Mark

Evaluate: $\int \frac{x-4}{(x-2)^3} \cdot e^x d x$

Answer

$\text { (c) }: \int \frac{x-4}{(x-2)^3} \cdot e^x d x=\int\left[\frac{x-2}{(x-2)^3}-\frac{2}{(x-2)^3}\right] e^x d x$
$=\int\left[\frac{1}{(x-2)^2}-\frac{2}{(x-2)^3}\right] e^x d x=\frac{e^x}{(x-2)^2}+C$
${\left[\because \int\left[f(x)+f^{\prime}(x)\right] e^x d x=e^x f(x)+C\right]}$

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MCQ 611 Mark

Value of $\int \cos 2 x d x$ is -

A
$-\frac{\sin 2 x}{2}+C$
✓
$\frac{\sin 2 x}{2}+C$
C
$-\sin 2 x+C$
D
$\sin 2 x+C$

Answer

Correct option: B.

$\frac{\sin 2 x}{2}+C$

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MATHS M.C.Q (1 Marks)